Question

The following equation involves a trigonometric equation in quadratic form. Solve the equation on the interval [0,2π).

Answer 1

Answer:

$\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$

Step-by-step explanation:

Given the quadratic equation as:

$2sin^2x+sinx=1\\OR\\2sin^2x+sinx-1=0$

Let us put $sinx=y$ for simplicity of the equation:

Now, the equation becomes:

$2y^2+y-1=0$

Now, let us try to solve the quadratic equation:

$\Rightarrow 2y^2+2y-y-1=0\\\Rightarrow 2y(y+1)-1(y+1)=0\\\Rightarrow (2y-1)(y+1)=0\\\Rightarrow 2y-1 = 0, y+1 = 0\\\Rightarrow y = \dfrac{1}{2}, y = -1$

So, the solution to the given trigonometric quadratic equation is:

$sinx = \dfrac{1}{2}$

and

$sinx=-1$

Let us try to find the values of $x$ in the interval $[0, 2\pi)$.

$sin\theta$ can have a value equal to $\frac{1}{2}$ in 1st and 2nd quadrant.

So, $x$ can be

$30^\circ, 150^\circ\\OR\\\dfrac{\pi}{6}, \dfrac{5\pi}{6}$

For $sinx=-1$,

$x = 270^\circ\ or\ \dfrac{3 \pi}{2}$

So, the answer is:

$\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$